576: Simplex Interior of Affine Simplex Is Open on Affine Simplex with Canonical Topology

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description/proof of that simplex interior of affine simplex is open on affine simplex with canonical topology

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Topics

About: vectors space
About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of simplex interior of affine simplex.
• The reader knows a definition of subspace topology of subset of topological space.
• The reader knows a definition of canonical $C^{\mathrm{\infty }}$ atlas for finite-dimensional real vectors space.
• The reader admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V$: $\in \{\text{ the }d\text{ -dimensional real vectors spaces }\}$ with the canonical topology
$\{p_0,...,p_n\}$: $\subseteq V$, $\in \{\text{ the affine-independent sets of base points on }V\}$
$[p_0,...,p_n]$: $=\text{ the affine simplex }$ as the topological subspace of $V$
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Statements:
$[p_0,...,p_n{]}^{\circ }\in \{\text{ the open subsets of }[p_0,...,p_n]\}$.
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2: Natural Language Description

For any $d$-dimensional real vectors space, $V$, with the canonical topology, any affine-independent set of base points on $V$, $\{p_0,...,p_n\}\subseteq V$, and the affine simplex, $[p_0,...,p_n]$, as the topological subspace of $V$, the simplex interior, $[p_0,...,p_n{]}^{\circ }$ is an open subset of $[p_0,...,p_n]$.

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3: Proof

$[p_0,...,p_n{]}^{\circ }=[p_0,...,p_n]\setminus {\cup }_{j\in J}{F}_j$, where $\{F_j|j\in J\}$ is the set of the proper faces of $[p_0,...,p_n]$.

When $n=0$, $[p_0{]}^{\circ }=[p_0]$, which is open on $[p_0]$.

Let us suppose that $1\le n$.

Let $(V,\varphi )$ be the canonical chart with respect to any basis of $V$, $\{p_1-p_0,...,p_n-p_0,b_{n+1},...,b_d\}\subseteq V$.

For any $p^{\prime }\in [p_0,...,p_n]$, $p^{\prime }=\sum _{j\in \{0,...,n\}}{t}^{\prime j}{p}_j=\sum _{j\in \{0,...,n\}}{t}^{\prime j}(p_j-p_0)+\sum _{j\in \{0,...,n\}}{t}^{\prime j}{p}_0=\sum _{j\in \{1,...,n\}}{t}^{\prime j}(p_j-p_0)+p_0$, which means that $\varphi (p^{\prime })=(t^{\prime 1}+p_0^1,...,t^{\prime n}+p_0^n,p_0^{n+1},...,p_0^d)$, where $\varphi (p_0)=(p_0^1,...,p_0^d)$.

Let $p\in [p_0,...,p_n]\setminus {\cup }_{j\in J}{F}_j$ be any. $\varphi (p)=(t^1+p_0^1,...,t^n+p_0^n,p_0^{n+1},...,p_0^d)$.

Let us take $\delta$ such that $0<\delta <min(t^1,...,t^n,(1-\sum _{j\in \{1,...,n\}}{t}^j)/n,1-t^1,...,1-t^n,(\sum _{j\in \{1,...,n\}}{t}^j)/n)$: $0<min(t^1,...,t^n,(1-(\sum _{j\in \{1,...,n\}}{t}^j))/n,1-t^1,...,1-t^n,(\sum _{j\in \{1,...,n\}}{t}^j)/n)$, because $p$ is not on any proper face of $[p_0,...,p_n]$ (the condition of being on a proper face is that $t^j=0$ for a $j\in \{0,...,n\}$, while $t^0=1-(\sum _{j\in \{1,...,n\}}{t}^j)$, and $t^j=1$ implies that $t^k=0$ for $k\ne j$). Let us take $U_{\varphi (p)}=(t^1+p_0^1-\delta ,t^1+p_0^1+\delta )\times ...\times (t^n+p_0^n-\delta ,t^n+p_0^n+\delta )\times (p_0^{n+1}-\delta ,p_0^{n+1}+\delta )\times ...\times (p_0^d-\delta ,p_0^d+\delta )\subseteq {\mathbb{R}}^d$, an open neighborhood of $\varphi (p)$ on ${\mathbb{R}}^d$.

$U_{\varphi (p)}\cap \varphi ([p_0,...,p_n])$, an open neighborhood of $\varphi (p)$ on $\varphi ([p_0,...,p_n])$, is $(t^1+p_0^1-\delta ,t^1+p_0^1+\delta )\times ...\times (t^n+p_0^n-\delta ,t^n+p_0^n+\delta )\times \{p_0^{n+1}\}\times ...\times \{p_0^d\}$, because $\delta$ has been chosen such that $0<t^j-\delta <t^j+\delta <1$ and $0<1-(\sum _{j\in \{1,...,n\}}(t^j+\delta ))=1-(\sum _{j\in \{1,...,n\}}{t}^j)-n\delta <1-(\sum _{j\in \{1,...,n\}}(t^j-\delta ))=1-(\sum _{j\in \{1,...,n\}}{t}^j)+n\delta <1$: for any $v=(t^{\prime 1}+p_0^1,...,t^{\prime n}+p_0^n,p_0^{n+1},p_0^d)\in (t^1+p_0^1-\delta ,t^1+p_0^1+\delta )\times ...\times (t^n+p_0^n-\delta ,t^n+p_0^n+\delta )\times \{p_0^{n+1}\}\times ...\times \{p_0^d\}$, $t^j+p_0^j-\delta <t^{\prime j}+p_0^j<t^j+p_0^j+\delta$, which implies that $0<t^j-\delta <t^{\prime j}<t^j+\delta <1$ and $\sum _{j\in \{1,...,n\}}{t}^j-n\delta <\sum _{j\in \{1,...,n\}}{t}^{\prime j}<\sum _{j\in \{1,...,n\}}{t}^j+n\delta$, which implies that $0<1-\sum _{j\in \{1,...,n\}}{t}^j-n\delta <1-\sum _{j\in \{1,...,n\}}{t}^{\prime j}<1-\sum _{j\in \{1,...,n\}}{t}^j+n\delta <1$, and so, there is $p^{\prime }=\sum _{j\in \{1,...,n\}}{t}^{\prime j}(p_j-p_0)+p_0\in [p_0,...,p_n]$ such that $\varphi (p^{\prime })=v$. $U_{\varphi (p)}\cap \varphi ([p_0,...,p_n])\subseteq \varphi ([p_0,...,p_n]\setminus {\cup }_{j\in J}{F}_j)$, because being on an $F_j$ means that $t^{\prime k}=0$ for a $k\in \{0,...,n\}$, but we already know that $0<t^{\prime j}<1$ for $j\in \{0,...,n\}$.

As $\varphi$ is a homeomorphism, ${\varphi }^{-1}(U_{\varphi (p)}\cap \varphi ([p_0,...,p_n]))$ is an open neighborhood of $p$ on $[p_0,...,p_n]$ such that ${\varphi }^{-1}(U_{\varphi (p)}\cap \varphi ([p_0,...,p_n]))\subseteq [p_0,...,p_n]\setminus {\cup }_{j\in J}{F}_j$. So, $[p_0,...,p_n]\setminus {\cup }_{j\in J}{F}_j$ is open on $[p_0,...,p_n]$, by the local criterion for openness.

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References

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